I this presentation me and group had worked on Guass's Law. It includes Introduction, Statements, Understandings and Applications in detail..

3.  Gauss’s law was formulated by German scientist
Carl Friedrich Gauss in 1835, but was not published
until 1867.
 Gauss's law, also known as Gauss's flux theorem and
Maxwell’s first equation.
 Gauss’s law is relating the distribution of electric
charge to the resulting electric field.
 Gauss’ Law allows us to find electric fields without
needing to integrate
 The electric field of a given charge distribution can in principle be
calculated using Gauss’ law.
Carl Friedrich

4. The total electric flux through a closed surface
is proportional to the enclosed charge
Where:
E = Electric Field
dA = Area Vector
Σq = Sum of all charges
ε0 = Permittivity of free space
Value of Permittivity of free space ε0=8.8542x10-12 C2/(N m2)

5.  Gauss’s Law is just a flux calculation
 Gauss’s Law only applies to closed surfaces called Gaussian Surface.
 Gauss’s Law directly relates electric flux to the charge distribution that
creates it.
 Gauss's law can be used to derive Coulomb's law, and vice versa.

6. Electric Field Intensity due to Spherical
Charge Distribution
Consider a charge “q” uniformly distributed on a closed sphere of radius “a”
as shown in figure

7. When point “p” Lies Outside the
Charge Sphere
 E . dA = q / e0
 E . dA = E dA = E A
 A = 4 r2
 E A = E 4 r2 = q / e0
 k = 1 / 4 0
 0 = permittivity
 0 = 8.85x10-12 C2/Nm2
 2
04
1
r
q
E

8. When point “p” Lies Inside the
Charge Sphere
From figure it is clear that there is no charge enclosed by Gaussian surface
therefore flux passes through it is zero, it means electric field intensity is also
zero.
 E 4 r2 = q / e0
 For Gaussian surface q=0
 E 4 r2 = 0 / e0
 E = 0 / 4 r2
 E = 0

9. When point “p” Lies on the Surface
of Charge Sphere
 E 4 r2 = q / e0
 E 4 a2 = q / e0 (where r = a)
 σ= q / ∆A
 q = σ . ∆A
 q= σ . 4 a2
 E 4 a2 = σ . 4 a2 / e0
 E = σ / e0

10. Electric Field Intensity due to Infinite
Sheet of Charge
 Select Gauss surface In this case
a cylindrical pillbox
 Calculate the flu of the electric
field through the Gauss surface
= 2 E A
 Equate = qencl/ 0 2EA = qencl/ 0
 Solve for E: E = qencl / 2 A 0 =
/ 2 0 (with = qencl / A)

11. Electric Field Intensity due to Oppositely
Charge Sheet
 E = E1 + E2
 E = σ / 2 e0 + σ / 2 e0
 E = 2 σ / 2 e0
 E = σ / e0